For the equilibirum, `PCl_(5)(g) hArr PCl_(3)(g) + Cl_(2)(g)` at 298 K, `K = 1.8 xx 10^(-7)`. Calculate `Delta G^(@)` for the reaction `(R = 8.314 JK^(-1) mol^(-1))`.

For the reaction : PCl_(5)(g) rarr PCl_(3)(g) + Cl_(2)(g)

Write the Kp expression for PCl_(5(g))hArr PCl_(3(g)) + Cl_(2(g))

The equilibrium constant at 25^(@)C for the process CO^(3+)(aq) + 6NH_(3)(aq) hArr [CO(NH_(3))_(6)]^(3)(aq) is 2.5 xx 10^(6) . Calculate the value of Delta G^(@) at 25^(@)C . (R = 8.314 JK^(-1) mol^(-1)) . In which direction is the reaction spontaneous under standard conditions ?

If equilibrium constant K is 10^(3) , the Delta G^(theta) for the reaction at 300 K is (assume R = 8.314 JK^(-1) mol^(-1) ) :

For the reaction, 2NOCl(g)hArr2NO(g)+Cl_(2)(g) the value of K_(c)=3.75xx10^(-6) at 1069K. Calculate K_(p) .

For the reaction N_(2)+3H_(2)hArr2NH_(3) at 773K, the value of K_(p)=1.4xx10^(-15) . Calculate K_(c) (Given R = 8.314 JK^(-1)mol^(-1) ).

At a given temperature, the equilibrium constant for the reaction : PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) is 2.4xx10^(-3) . At the same temperature, the equilibrium constant for the reaction : PCl_(3)(g)+Cl_(2)(g) harr PCl_(5)(g) is :

A plot of (1)/(T) Vsk for a reaction gives the slope - 1 xx 10^(4)K . The energy of activation for the reaction is (Given R = 8.314 JK^(-1) " mol"^(-1) )

DeltaH for the reaction at 298 K CO(g)+1//2O_(2)(g) is 282.85 KJ mol^(-1) . Calculate DeltaU of the reaction.